STADs - Shear force and bending in simple beams

The Fact File

General definition of shearing force
The shearing force at any section in a horizontal beam is the algebraic sum of all the vertical forces to the left, or to the right, of that section.
General definition of bending moment
The bending moment at any section in a beam is the algebraic sum of the moments about that section of all the forces to the left, or to the right, of that section.
Shearing force diagram
A shearing force diagram is a graph showing the variation of the shearing force along the length of the beam or other structural member.
Shearing force diagrams resulting from the action of concentrated loads on horizontal beams consist of horizontal straight lines with vertical jumps at load positions.
Shearing force diagrams resulting from the actions of uniformly distributed loads on horizontal beams consist of sloping straight lines.
Bending moment diagram
A bending moment diagram is a graph showing the variation of the bending moment along the length of a beam or other structural member.
Bending moment diagrams resulting from the action of concentrated point loads consist of straight lines with changes of direction at the points of application of the loads.
Bending moment diagrams resulting from the action of uniformly distributed loads consist of parabolic curves.
At the point of application of a couple to a horizontal beam, there will be a vertical step in the bending moment diagram.
The bending moment at an internal pin or a pinned support is zero.
Points of contraflexure
The direction of curvature changes at a point of contraflexure - that is, at a point where the bending moment is zero.
Standard formulae for maximum bending moment
If a single concentrated vertical load W acts at the mid-point of a simply supported beam of span L, then the maximum bending moment is at the point of application of the load and is of magnitude WL/4.
If a uniformly distributed load w per unit length acts along the entire length of a simply supported beam of span L, then the maximum bending moment is at mid-span and is of magnitude wL²/8.

Symbols, units and sign conventions

H	=	horizontal reactions at the support (kN)
V	=	vertical reactions at the support (kN)
B.M.	=	bending moment (kNm)
S.F.	=	shear force (kN)

The shearing force at a section in a beam is taken to be positive if the resultant vertical loading on the part of the beam to the left of that section tends to move that part of the beam vertically upwards.
The bending moment in a beam is taken to be positive if it causes the beam to sag.
Bending moment diagrams are plotted such that the positive bending moments are on the tensile side of the beam or other structural member.

Video Tutorials

Study Notes

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T301.1) that graphically illustrates the beam and the external forces acting on it.

Figure T301.1 T1 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that there are no horizontal loads applied to the beam, therefore:

Vertical Reactions

Consider the reaction at "A", taking moments about "B" and assuming that any loads or forces act in a clockwise direction about B are positive:

For the more general case, where there is a universally distributed load (UDL) across the full span of the beam:

Shear Force Diagram

Bending Moment

The maximum bending moment, for a beam with a universally distributed load along its full length, is at the mid-span point. It is also useful to remember that where the shear force is zero coincides with the maximum bending in the beam.

The position of the maximum bending moment (X), measured from A:

N.B. Bending moments that cause a beam to sag are taken as being positive.

Taking moments about the centre of the beam where the bending moment is at its maximum gives:

Bending Moment Diagram

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T301.1) that graphically illustrates the beam and the external forces acting on it.

Figure T301.1 T1 - Free Body Diagram.

Static Equilibrium.

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:

Vertical Reactions

For a simply supported beam, where the only load is a universally distributed load (a constant load along the full length of the beam), the vertical reactions at the supports can be calculated one of two ways.

By taking moments about one support to determine the other, or
By simply recognising that the load will be equally distributed between the two vertical supports.

Consider the reaction at "B". This can be calculated by taking moments about "A".

Figure T301.1 illustrates the concept of "positive" and "negative" rotation of a load or reaction when taking moments about A. The UD load acts in a clockwise direction about A and is considered "POSITIVE" ("+ve"), whereas the vertical reaction Vb is acting in an upwards direction tending to rotate anti-clockwise about A and is therefore considered to be "NEGATIVE" ("-ve").

Figure T301.1 T1 - Positive & Negative Rotation About "A"

For the reaction at B, using the third equation of equilibrium and taking moments about A, we can say:

We now formulate an expression that equates the moments generated by the UD load (w) and the vertical reaction (Vb), when acting about A, to zero:

Substituting in the tutorial data gives:

Finally, solving for the reaction R_B:

For the vertical reaction at A, we adopt the first equation of equilibrium:

Equating the vertical actions (loads and reactions) to zero, we can say:

NOTE: The direction of the arrows indicating the load / reactions. Vertically upwards = POSITIVE, vertically downward = NEGATIVE.

Substituting in the tutorial data and the calculated value of the reaction at B, gives:

Finally, solving for the reaction R_A:

The method above outlines the way in which to calculate the reactions from first principles. For the more general case, where there is a universally distributed load (UDL) across the full span of the beam the vertical reaction at either A or B is simply a case of the total load divided by 2, i.e:

Shear Force

A Shear Force Diagram is a means of graphically illustrating the change in shear force along the length of the beam. The horizontal axis represents the length of the beam, and the vertical axis represents the shear force.

The first step in constructing a shear force diagram is to draw a horizontal line that represents the beam. It is then a case of plotting the reaction at A and then incrementally along the beam plotting the change in shear force - Figure 301.2 illustrates this principle.

Figure T301.2 Construction of the Shear Force Diagram

Figure 301.1 (i) to (vi) shows the incremental construction of the Shear Force Diagram.

(i) - A vertical line is drawn that represents the reaction at A. If the reaction acts up, then the line representing this is drawn up. NOTE: The lines making up the shear force diagram are drawn in the same direction as the actions they represent.
(ii) - @ 1m along the beam we consider everything to the left of the 1m line. From 0m to 1m the only change in the actions is 1m worth of the UD load w, acting down. Therefore the value of the shear force at 1m = Ra - w.
This same principle is carried on for the full length of the beam. Therefore @ 5m (v) - along the beam we consider everything to the left of the 5m line. From 0m to 5m the only change in the actions is 5m worth of the UD load w, acting down. Therefore the value of the shear force at 5m = Ra - 5w.
At the right hand end of the beam (vi) the shear force at 6m = Ra - 6w. At this point the vertical (upwards) reaction Rb acts. We therefore draw a vertical line representing Rb. As this is the end of the beam, and to maintain static equilibrium, the shear force at 6m (Ra - 6w) should equal the vertical reaction Rb, such that:

Solution Shear Force Diagram

Bending Moment

The Bending Moment Diagram is a graph that shows the variation in bending along a beam. For its construction we must consider "HOGGING" and "SAGGING" in a beam and how these produce either "POSITIVE" or "NEGATIVE" bending moments.

NOTE: The calculations associated with bending moments have nothing to do with the direction of rotation, only if the applied action results in positive or negative bending.

Figure T301.3 Positive and Negative Bending Moments

The maximum bending moment, for a beam with a universally distributed load along its full length, is at the mid-span point.

It is also useful to remember that where the shear force is zero coincides with the maximum bending in the beam.

The position of the maximum bending moment (X), measured from A:

The procedure for taking moments about the centre (or any point on a beam) is as follows:

Step 1		"CUT": The beam at the point about which you are taking moments. In this case we cut the beam in the center where the bending moment is the maximum.
Step 2		"BIN": Ignore one half of the beam and fix the half you are interested in. In this case we are looking at the left hand end. This section has been given a notional fixed support. "ONE END IN": Working from the end of the beam to the fixed support, consider each action in turn and formulate expressions for the moments they generate.
Step 3		Reaction Ra: The vertically upwards reaction at A causes TENSION in the bottom of the beam. It is therefore a SAGGING moment and POSITIVE. Taking moments about the fixed end gives:
Step 4		UDL - w: The UDL acts vertically downwards causing TENSION in the top of the beam. It is therefore a HOGGING moment and NEGATIVE. Taking moments about the fixed end gives:
A useful mnemonic to remember: "CUT, BIN, ONE END IN"

Putting the above two expressions together gives a single expression for the bending moment at the mid-span of the beam.

Substituting in the design data:

Solving, gives the final mid-span moment:

For the general case where a simply supported beam has a UDL across the full span the bending moment can be calculated as follows (you should remember this!):

Bending Moment Diagram

It can be shown that the shape of the bending moment diagram for a simply supported beam with a UD load across the full length is that of a parabola:

Figure T301.4

Consider the beam portion shown in Figure T301.4.

If we take moments about an arbitrary point, X from the support at A, we can derive an expression for the bending moment at that point.

"CUT": Cut the beam at X.

"BIN": Ignore anything to the right of X.

"ONE END IN": Working from the end of the beam, derive expressions for the bending moments.

For the reaction at A (SAGGING):

For the UD load (HOGGING):

Combining these gives an equation for the Bending Moment at X and also the equation of a parabola.

Solution Bending Moment Diagram

NOTE: When construction the bending moment diagram:

The bending moment is ALWAYS drawn on the tension side of the beam.
POSITIVE, SAGGING bending moments where the bottom of the beam is in tension are drawn below the line.
NEGATIVE, HOGGING bending moments where the top of the beam is in tension are drawn above the line.

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T302.1) that graphically illustrates the beam and the external forces acting on it.

Figure T302.1 T2 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that the applied point load acts vertically, with no horizontal component, it can be said that:

Therefore the horizontal reactions at both A and B = 0.

Vertical Reactions

Consider the reaction at "A", taking moments about "B" and assuming that any loads or forces acting in a clockwise direction about B are positive:

For the more general case, where where the point load acts in the centre of the beam the reactions can be calculated as follows:

Shear Force Diagram

Bending Moment

The maximum bending moment, for a beam with a single point load, is at the position of the point load. It is also useful to remember that where the shear force is zero coincides with the maximum bending in the beam.

The position of the maximum bending moment (X), measured from A:

N.B. Bending moments that cause a beam to sag are taken as being positive.

Taking moments about the centre of the beam where the bending moment is at its maximum gives:

Bending Moment Diagram

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T302.1) that graphically illustrates the beam and the external forces acting on it.

Figure T302.1 T2 - Free Body Diagram.

Static Equilibrium

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:

Vertical Reactions

For a simply supported beam, subjected to a point load at a distance L2 from the support, the vertical reactions at the supports can be calculated by taking moments about one support to determine the vertical reaction at the other i.e. taking moments about A, will give the vertical reaction at B, whereas taking moments about B will give the vertical reaction at A.

Consider the reaction at "A". This can be calculated by taking moments about "B". For this, it is usual to assume that any loads or forces acting in a clockwise direction about the point of interest are positive, i.e. any loads acting clockwise about "B" will be treated as positive".

Figure T301.1 illustrates the concept of "positive" and "negative" rotation of a load or reaction when taking moments about B. The Point Load (PL) load acts in a anti-clockwise direction about B and is considered "NEGATIVE" ("-ve"), whereas the vertical reaction Va is acting in an upwards direction tending to rotate clockwise about B and is therefore considered to be "POSITIVE" ("+ve").

Figure T302.1 T1 - Positive & Negative Rotation About "B"

For the reaction at A, using the third equation of equilibrium and taking moments about B, we can say:

We now formulate an expression that equates the moments generated by the Point Load (PL) and the vertical reaction (Va), when acting about B, to zero:

Substituting in the tutorial data gives:

Finally, solving for the reaction R_A:

For static equilibrium, and to satisfy one of the equations of equilibrium, the sum of the vertical forces must equate to zero. If this condition is not satisfied then the beam becomes unstable. To meet this requirement it is simply a case of equating the applied loads to the reactions i.e the sum of the reactions must equal the sum of the point loads. This can be achieved in two ways:

By simply equating the sum of the vertical reactions and forces to zero.
By taking moments about A to give the vertical reaction at B. For this we usual to assume that any loads acting clockwise about "A" will be treated as positive".

The method above outlines the way in which to calculate the reactions from first principles. For the more general case, where there is a single point load at the mid-span point span of the beam the vertical reaction at either A or B is simply a case of the total point load divided by 2, i.e:

Shear Force

A Shear Force Diagram is a means of graphically illustrating the change in shear force along the length of the beam. The horizontal axis represents the length of the beam, and the vertical axis represents the shear force.

The first step in constructing a shear force diagram is to draw a horizontal line that represents the beam. It is then a case of plotting the reaction at A and then incrementally along the beam plotting the change in shear force - Figure 302.2 illustrates this principle.

Figure T302.2 Construction of the Shear Force Diagram

Figure 302.2 (i) to (ii) shows the incremental construction of the Shear Force Diagram.

(i) - A vertical line is drawn that represents the reaction at A. If the reaction acts up, then the line representing this is drawn up. NOTE: The lines making up the shear force diagram are drawn in the same direction as the actions they represent.
(ii) - From A, to just left of the Point Load there is no change to the actions being applied to the beam. As such there can be no change to the shear force as this would require a change in the loading condition. A horizontal line is drawn to represent the constant shear force along this portion of beam.
(iii) - At the Point Load - At the position where the point load is applied there is a vertical change to the shear force that equals both the magnitude of the point load and the direction in which it is applied. In this case the point load acts down, therefore the line representing this is drawn down.
To the right of the Point Load there is no change to the applied loads until the end of the beam, so therefore there is no change in the shear force - a horizontal line is drawn to reflect this.
The vertical (upwards) reaction Rb acts to the right hand end of the beam. It's magnitude, to maintain static equilibrium, must equate to: Rb = Ra - PL. A vertical line in the direction of Rb is drawn to finalise the shear force diagram.

Shear Force Diagram

Bending Moment

The Bending Moment Diagram is a graph that shows the variation in bending along a beam. For its construction we must consider "HOGGING" and "SAGGING" in a beam and how these produce either "POSITIVE" or "NEGATIVE" bending moments.

NOTE: The calculations associated with bending moments have nothing to do with the direction of rotation, only if the applied action results in positive or negative bending.

Figure T302.3 Positive and Negative Bending Moments

The maximum bending moment, for a beam with a single point load, is at the position of the point load. It is also useful to remember that where the shear force is zero coincides with the maximum bending in the beam.

The position of the maximum bending moment (X), measured from A:

The procedure for taking moments about the centre (or any point on a beam) is as follows:

Step 1		"CUT": The beam at the point about which you are taking moments. In this case we cut the beam at the position of the point load, where the bending moment is the maximum. "BIN": Ignore one half of the beam and fix the half you are interested in. In this case we are looking at the left hand end. This section has been given a notional fixed support at the "cut".
Step 2		"ONE END IN": Working from the end of the beam to the fixed support, consider each action in turn and formulate expressions for the moments they generate. Reaction Ra: The vertically upwards reaction at A causes TENSION in the bottom of the beam. It is therefore a SAGGING moment and POSITIVE. Taking moments about the fixed end gives:
A useful mnemonic to remember: "CUT, BIN, ONE END IN"

Taking moments about the the position of the point load, where the bending moment is at its maximum gives:

Substituting in the design data:

Solving, gives the final maximum bending moment:

Bending Moment Diagram

NOTE: When construction the bending moment diagram:

The bending moment is ALWAYS drawn on the tension side of the beam.
POSITIVE, SAGGING bending moments where the bottom of the beam is in tension are drawn below the line.
NEGATIVE, HOGGING bending moments where the top of the beam is in tension are drawn above the line.

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T303.1) that graphically illustrates the beam and the external forces acting on it.

Figure T303.1 T3 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that the applied loads (both the UDL and the Point Load) act vertically, with no horizontal components, it can be said that:

Vertical Reactions

Consider the reaction at "B", taking moments about "A" and assuming that any loads or forces acting in a clockwise direction about A are positive:

Substituting the tutorial data gives:

For the vertical reaction at A:

Shear Force Diagram

Bending Moments

The position (X) along the beam where the bending moment is a maximum:

Maximum span bending moment @ X:

Bending moment at B:

Bending Moment Diagram

The point of contraflexure (x̄) from A:

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T303.1) that graphically illustrates the beam and the external forces acting on it.

Figure T303.1 T3 - Free Body Diagram.

Static Equilibrium

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:

Vertical Reactions

For a simply supported beam, albeit one with a cantilever to one end, where the loads are acting in such a manner that the vertical reactions at the supports are not equal then the first step to calculating the vertical reactions is to take moments about EITHER support to determine the other. Once this is done, the second support reaction can be determined by simple statics, recognising that the sum of the vertical forces equates to zero.

As we can take moments about either support it would seem logical that we choose the least complex reaction to calculate first.

Looking at the beam diagram we can see the following:

The full UD load and the point load are both acting clockwise about A.
The UD load to the left of the support at B acts anti-clockwise about B
The UD load to the right of B acts clockwise about B.
The point load to the end of the cantilever acts clockwise about B.

In this instance it is therefore easiest to firstly calculate the vertical reaction at B, by taking moments about A.

For this, it is usual to assume that any loads or forces acting in a clockwise direction about the point of interest are positive, i.e. any loads acting clockwise about "A" will be treated as positive".

Reaction at B

The first step to calculate the reaction at B is to formulate an expression that considers the moments generated by the individual forces action about A, and equating these to zero. This satisfies one of the three equations of equilibrium, i.e. ΣM = 0

There are three forces acting on the beam that must be considered when calculating the reaction at B: The following three expressions consider the individual force and the moment it generated when acting about A - Remember that clockwise forces are assumed to be positive, and the a moment is defined as a force times a lever arm.

The UD load to the full length of the beam (acting clockwise about A, ∴ considered positive):
The point load to the end of the cantilever (acting clockwise about A, ∴ considered positive):
The vertical reaction at B (acting anti-clockwise about A, ∴ considered negative):

Equating these three expressions to zero gives a single expression for the sum of the moments about support A.

It's now a simple case of substituting into the above expression the relevant values for the beam spans and loads. This can then be solved to give the reaction at B.

Simplifying gives:

Reaction at A

The reactions at A can now be calculated from simple statics, recognising that the sum of the vertical forces equates to zero i.e. ΣV = 0.

To demonstrate the point made earlier, that you can take moments about either support to determine the vertical reaction of the other support, the following will take moments about B to determine the vertical reaction at A.

Shear Force Diagram

Bending Moments

Maximum Moment in Span

The maximum bending moment in the beam between A & B occurs at a point along the beam where the shear force is zero.

The position along the beam (X) where the shear force is zero, and the bending moment is a maximum, is calculated by taking the reaction at A (R_A) and dividing it by the magnitude of the UD load "w", such that:

Taking moments about the point X will give the maximum span bending moment (Remember: POSITIVE bending gives tension on the bottom of the beam & NEGATIVE bending gives tension on the top of the beam):

Bending Moment @ B

The magnitude of the bending moment at any point along a beam can be calculated by taking moments about the point of interest. Therefore at B, the bending moment can be calculated as follows:

Bending Moment Diagram

NOTE: When construction the bending moment diagram:

The bending moment is ALWAYS drawn on the tension side of the beam.
POSITIVE, SAGGING bending moments where the bottom of the beam is in tension are drawn below the line.
NEGATIVE, HOGGING bending moments where the top of the beam is in tension are drawn above the line.

Point of Contraflexure

The point of contraflexure occurs at the point on a beam where the bending moment changes sign from positive bending (sagging) to negative bending (hogging). At this point (x̄) the bending moment is zero, such that:

This quadratic equation (in the form of: ax² + bx + c = 0) is then solved, giving x̄ as the position of the point of contraflexure:

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T304.1) that graphically illustrates the beam and the external forces acting on it.

Figure T304.1 T4 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that the applied loads (both the UDL and the Point Load) act vertically, with no horizontal components, it can be said that:

Vertical Reactions

Consider the reaction at "B", taking moments about "A" and assuming that any loads or forces acting in a clockwise direction about A are positive:

Substituting the tutorial data gives:

For the vertical reaction at A:

Shear Force Diagram

Bending Moments

Bending moment at 1:

Bending moment at 2:

Bending moment at 3:

Bending moment at 4:

Bending moment at C:

Bending Moment Diagram

Point of Contraflexure

Point of contraflexure, x̄ from A

Substituting in the tutorial data and solving for x̄ gives:

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T304.1) that graphically illustrates the beam and the external forces acting on it.

Figure T304.1 T4 - Free Body Diagram.

Static Equilibrium

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:

Vertical Reactions

For a simply supported beam, albeit one with a cantilever to one end, where the loads are acting in such a manner that the vertical reactions at the supports are not equal then the first step to calculating the vertical reactions is to take moments about EITHER support to determine the other. Once this is done, the second support reaction can be determined by simple statics, recognising that the sum of the vertical forces equates to zero.

As we can take moments about either support it would seem logical that we choose the least complex reaction to calculate first.

Looking at the beam diagram we can see the following:

The partially distributed load and both the point loads are both acting clockwise about A.
The extents of the partially distributed load is determined by the dimensions L3 and L4, as such, it could be positioned on the beam so that it runs over the top of the support at B. This would mean that the first part of the load (L3 - B) would act anti-clockwise about B, and the remainder that is on the cantilevered section of the beam (from B to L4) would act clockwise about B.
The positioning of the point loads is such that P1 is always between A & B, therefore acting anti-clockwise about, whereas, P2 is positioned on the cantilever and acts clockwise about B.

In this instance it is therefore easiest to firstly calculate the vertical reaction at B, by taking moments about A.

For this, it is usual to assume that any loads or forces acting in a clockwise direction about the point of interest are positive, i.e. any loads acting clockwise about "A" will be treated as positive".

The first step to calculate the reaction at B is to formulate an expression that considers the moments generated by the individual forces action about A, and equating these to zero. This satisfies one of the three equations of equilibrium, i.e. ΣM = 0

There are four forces acting on the beam that must be considered when calculating the reaction at B: The following three expressions consider the individual force and the moment it generated when acting about A - Remember that clockwise forces are assumed to be positive, and the a moment is defined as a force times a lever arm.

The partially distributed load to a portion of the beam/cantilever (acting clockwise about A, ∴ considered positive):
The point load positioned at any point on the cantilever (acting clockwise about A, ∴ considered positive):
The point load to the main span between A & B (acting clockwise about A, ∴ considered positive):
The vertical reaction at B (acting anti-clockwise about A, ∴ considered negative):

Equating these three expressions to zero gives a single expression for the sum of the moments about support A.

It's now a simple case of substituting into the above expression the relevant values for the beam spans and loads. This can then be solved to give the reaction at B.

Simplifying gives:

The reactions at A can now be calculated from simple statics, recognising that the sum of the vertical forces equates to zero i.e. ΣV = 0.

To demonstrate the point made earlier, that you can take moments about either support to determine the vertical reaction of the other support, the following will take moments about B to determine the vertical reaction at A.

Shear Force Diagram

Bending Moments

Bending moment at 1:

Bending moment at 2:

Bending moment at 3:

Bending moment at 4:

Bending moment at C:

Bending Moment Diagram

Point of Contraflexure

Substituting in the tutorial data and solving for x̄ gives:

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T305.1) that graphically illustrates the beam and the external forces acting on it.

Figure T305.1 T5 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that the applied point load P2 acts an angle θ° from the horizontal it has both a vertical and horizontal component. The first step is to resolve the inclined load into its vertical and horizontal components. Once done the sum of the horizontal force components will give the horizontal reaction (Ha) at A.

The vertical and horizontal components of P2 are as follows:

When calculating the horizontal reaction H_A it is important that the direction of the loads are taken in the global axis (Right Hand Rule). In the global axis, loads acting from LEFT to RIGHT in the direction of the X axis are taken to be POSITIVE.

Vertical Reactions

Consider the reaction at "B", taking moments about "A" and assuming that any loads or forces acting in a clockwise direction about A are positive:

Substituting the tutorial data gives:

For the vertical reaction at A:

Shear Force Diagram

Bending Moments

Maximum span bending (X from A) occurs where the shear force is zero:

Bending moment at Support A:

Bending moment at Support B:

Bending moment at 1 (start of PDL):

Bending moment at 2 (end of PDL):

Bending moment at 3 (@ P2)

Bending Moment Diagram

Point of Contraflexure

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T305.1) that graphically illustrates the beam and the external forces acting on it.

Figure T305.1 T5 - Free Body Diagram.

Static Equilibrium

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

Given that the applied point load P2 acts an angle θ° from the horizontal it has both a vertical and horizontal component. The first step is to resolve the inclined load into its vertical and horizontal components. Once done the sum of the horizontal force components will give the horizontal reaction (Ha) at A.

The vertical and horizontal components of P2 are as follows:

When calculating the horizontal reaction H_A it is important that the direction of the loads are taken in the global axis (Right Hand Rule). In the global axis, loads acting from LEFT to RIGHT in the direction of the X axis are taken to be POSITIVE.

Vertical Reactions

For a simply supported beam, albeit one with cantilevers to both ends, where the loads are acting in such a manner that the vertical reactions at the supports are not equal then the first step to calculating the vertical reactions is to take moments about EITHER support to determine the other. Once this is done, the second support reaction can be determined by simple statics, recognising that the sum of the vertical forces equates to zero.

As we can take moments about either support it would seem logical that we choose the least complex reaction to calculate first.

Looking at the beam diagram we can see the following:

The partially distributed load is arranged in such a manner that it can be distributed across the full length of the beam, or on only a portion of the beam. Depending on the arrangement it will generate both/either positive or negative bending about A & B.
Both the cantilevers have a point load to the their ends. P1 acts anti-clockwise about A, generating negative bending whereas P3 acts clockwise about B, but also generates negative bending.
The positioning of the point load P2 is such that it is always between A & B. It will act clockwise about A, and anti-clockwise about B.

In this case there is not really an easier reaction to calculate first! We will therefore consider the vertical reaction at B, by taking moments about A.

For this, it is usual to assume that any loads or forces acting in a clockwise direction about the point of interest are positive, i.e. any loads acting clockwise about "A" will be treated as positive".

The first step to calculate the reaction at B is to formulate an expression that considers the moments generated by the individual forces action about A, and equating these to zero. This satisfies one of the three equations of equilibrium, i.e. ΣM = 0

There are five forces acting on the beam that must be considered when calculating the reaction at B: The following expressions consider the individual forces and the moment it generated when acting about A - Remember that clockwise forces are assumed to be positive, and the a moment is defined as a force times a lever arm.

The point load P1, positioned at the end of the left hand cantilever (acting anti-clockwise about A, ∴ considered negative):
The partially distributed load to the left hand cantilever i.e. to the left of A (acting anti-clockwise about A, ∴ considered negative):
The partially distributed load to the main beam span between A and the right hand end of the beam (acting clockwise about A, ∴ considered positive):
The point load P2 (remember to use the vertical component of the point load), positioned on the beam between A & B (acting clockwise about A, ∴ considered positive):
The point load P3, positioned at the end of the right hand cantilever (acting clockwise about A, ∴ considered positive):
The vertical reaction at B (acting anti-clockwise about A, ∴ considered negative):

Equating these three expressions to zero gives a single expression for the sum of the moments about support A.

It's now a simple case of substituting into the above expression the relevant values for the beam spans and loads. This can then be solved to give the reaction at B.

Simplifying gives:

The reactions at A can now be calculated from simple statics, recognising that the sum of the vertical forces equates to zero i.e. ΣV = 0.

To demonstrate the point made earlier, that you can take moments about either support to determine the vertical reaction of the other support, the following will take moments about B to determine the vertical reaction at A.

Shear Force Diagram

Bending Moments

Maximum span bending (X from A) occurs where the shear force is zero.

Calculating the point of zero shear enables us to take moments of all the forces acting to either the left or the right (not both), about the point and equating these to zero.

REMEMBER: CUT: BIN: ONE END IN:

At each point where the moment is required:

CUT: the beam at the point of interest e.g. @ 1, the start of the UD load.
BIN: ignore everything either to the left or the right.
ONE END IN: working from the end of the beam to the cut, take moments about the cut for each applied force. With multiple forces formulate these as a single expression.

Maximum span bending (@ X from A):

Bending moment at Support A:

Bending moment at Support B:

Bending moment at 1 (start of PDL):

Bending moment at 2 (end of PDL):

Bending moment at 3 (@ P2)

Bending Moment Diagram

Point of Contraflexure

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T306.1) that graphically illustrates the beam and the external forces acting on it.

Figure T306.1 T6 - Free Body Diagram.

Static Equilibrium.

For a simply supported beam, where there are three unknowns that can be satisfied by the three equations of equilibrium it can be said that the beam is statically determinate such that:

Horizontal Reactions

Given that the applied loads (both the UDL and the Point Load) act vertically, with no horizontal components, it can be said that:

Vertical Reactions

Consider the reaction at "B", taking moments about "A" and assuming that any loads or forces acting in a clockwise direction about A are positive:

Substituting the tutorial data gives:

For the vertical reaction at A:

Shear Force Diagram

Bending Moments

Maximum span bending (X from A) occurs where the shear force is zero:

Figure T306.2 VPDL Similar Triangles.

With reference to Figure T306.2, the relationship between L3 and x̄ is that of similar triangles, such that:

The sum of the vertical loads to the left of the point of zero shear equal zero, such that:

Bending moment M_X, at X (m) from the support at A:

Substituting in the tutorial data gives:

Bending Moment Diagram

Beam span and loading.

Free Body Diagram

The first step is to draw a free body diagram (see Figure T306.1) that graphically illustrates the beam and the external forces acting on it.

Figure T306.1 T6 - Free Body Diagram.

Static Equilibrium

The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:

@ A - Vertical reaction, R_y, and @ B - Horizontal reaction, R_y
@ B - Vertical reaction, R_y

For the beam, this gives a total of three unknown forces, for which there are three equations of equilibrium:

Therefore: The beam can be said to be statically determinate.

Horizontal Reactions

The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:

Vertical Reactions

For a simply supported beam, where the loads are acting in such a manner that the vertical reactions at the supports are not equal then the first step to calculating the vertical reactions is to take moments about EITHER support to determine the other. Once this is done, the second support reaction can be determined by simple statics, recognising that the sum of the vertical forces equates to zero.

The procedure for calculating the vertical reactions is the same as has been discussed in the previous examples. The difference with this question is the configuration of the load.

The varying partially distributed load is essentially made up of two individual loads acting over the same portion length of the beam.

Figure T306.2 T6 - Varying Partially Distributed Load.

Considering Figure T306.2, this, when taking moments about A, illustrates the way a varying partially distributed load is split into its constituent parts.

Consider the reaction at B, by taking moments about A, and equating these to zero. This satisfies one of the three equations of equilibrium, i.e. ΣM = 0

The following expressions consider the individual forces and the moment they generated when acting about A - Remember that clockwise forces are assumed to be positive, and the a moment is defined as a force times a lever arm.

The partially distributed load - See Figure T306.2 (ii). The moment generated is total force acting over the length L3 multiplied by the lever arm (acting clockwise about A, ∴ considered positive):
The varying partially distributed load - see Figure T306.2 (iii). Again the moment generated is total force acting over the length L3 multiplied by the lever arm - Note, this acts from the centroid of the load (acting clockwise about A, ∴ considered positive):
The vertical reaction at B (acting anti-clockwise about A, ∴ considered negative):

Equating these three expressions to zero gives a single expression for the sum of the moments about support A.

It's now a simple case of substituting into the above expression the relevant values for the beam spans and loads. This can then be solved to give the reaction at B.

Simplifying gives:

The reactions at A can now be calculated from simple statics, recognising that the sum of the vertical forces equates to zero i.e. ΣV = 0.

Shear Force Diagram

Bending Moments

Maximum span bending (X from A) occurs where the shear force is zero.

From the shear force diagram we can see that zero shear occurs part way along the varying partially distributed load.

Figure T306.3 VPDL Similar Triangles.

With reference to Figure T306.3, the relationship between L3 and x̄ is that of similar triangles, such that:

The sum of the vertical loads to the left of the point of zero shear equal zero, such that:

Calculating the point of zero shear enables us to take moments of all the forces acting to either the left or the right (not both), about the point and equating these to zero.

REMEMBER: CUT: BIN: ONE END IN:

At each point where the moment is required:

CUT: the beam at the point of interest e.g. @ X, the point of zero shear where the span moment is a maximum.
BIN: ignore everything either to the left or the right.
ONE END IN: working from the end of the beam to the cut, take moments about the cut for each applied force. With multiple forces formulate these as a single expression.

Maximum span bending (@ X from A):

Note the use of "Y" for the magnitude of the triangular load at the cut (X from A). Y is given above as a multiple of x̄:

For the bending moment at 1, it is simple a case of taking moments of all the forces to the left of 1, i.e. the only force acting is the reaction at A. This gives a positive sagging moment, such that:

For the bending moment at 2, it is simple a case of taking moments of all the forces to the right of 2, i.e. the only force acting is the reaction at B. This gives a positive sagging moment, such that:

Shear Force and Bending Moments

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

Beam Data

Solution

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

Beam Data

Solution

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

Beam Data

Solution

Vertical Reactions

Bending Moment

Shear Force and Bending Moments

T1 : Simple beam with UDL.

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

T2 : Simple beam with PL.

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

T3 : Cantilever to right, UDL & PL.

Beam Data

Solution

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

T4 : Cantilever to right, PUDL & PL's.

Beam Data

Solution

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

T5 : Cantilever to left & right, PUDL & PL's.

Beam Data

Solution

Horizontal Reactions

Vertical Reactions

Bending Moment

Point of Contraflexure between A & B

T6 : Simple beam with VPDL.

Beam Data

Solution

Vertical Reactions

Bending Moment

T10 : Cantilever with PUDL & inclined PL.

T14 : 2 spans (int. pin), cantilever to right, 2 No PUDL's, 2 No. PL's & axial PL.